An object 5 millimeters high is located 15 millimeters in front of a plane mirro. How far from the mirror is the image located?

The focal length f of a plane mirror is infinity and 1/f = 0. The optical equation becomes

1/Do + I/Di = 1/f = 0

Do is the distance to the object and Di is the distance to the image (measured on the same side of the mirror).

Therefore Do = -Di The minus sign in front of Di means that the image appears on the other side of the mirror, an equal distance away from it.

wrong answer it has to be 5.0 millimeters or 7.5 millimeters or 15 millimeters or 30 millimeters

which one is it though?

An airplane reaches its takeoff speed of 60m/s in 30s starting from rest. the time it spends in going from 40m/s to 60m/s is

15 m I think, i have the same homework

To find the distance of the image from the mirror, we can use the mirror equation:

1/Do + 1/Di = 1/f

Since the mirror is a plane mirror with a focal length of infinity (f = ∞), the mirror equation simplifies to:

1/Do + 1/Di = 0

Given that the object distance (Do) is 15 millimeters and the height of the object is 5 millimeters, we can solve for the image distance (Di) by substituting these values into the mirror equation:

1/15 + 1/Di = 0

To solve for Di, we bring the equation to a common denominator:

(1 + 15/Di) / 15 = 0

Since the denominator cannot be zero, we can set the numerator equal to zero:

1 + 15/Di = 0

Now we solve for Di by isolating the variable:

15/Di = -1

Dividing both sides of the equation by 15, we get:

1/Di = -1/15

Inverting both sides of the equation, we get:

Di = -15

Since distance cannot be negative, we disregard the negative sign and take the absolute value:

Di = 15 millimeters

Therefore, the distance from the mirror to the image is 15 millimeters.

Regarding the second question about the airplane's takeoff speed, the information provided is not sufficient to determine the time spent going from 40 m/s to 60 m/s. Additional information about the acceleration or distance traveled during this time period is needed to calculate the time.